Question: You have found the following ages (in years) of all 4 seals at your local zoo: $ 3,\enspace 3,\enspace 20,\enspace 1$ What is the average age of the seals at your zoo? What is the variance? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 4 seals at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $4$ ages and divide by $4$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\mu} = \dfrac{3 + 3 + 20 + 1}{{4}} = {6.8\text{ years old}} $ Find the squared deviations from the mean for each seal. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $3$ years $-3.8$ years $14.44$ years $^2$ $3$ years $-3.8$ years $14.44$ years $^2$ $20$ years $13.2$ years $174.24$ years $^2$ $1$ year $-5.8$ years $33.64$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{14.44} + {14.44} + {174.24} + {33.64}} {{4}} $ $ {\sigma^2} = \dfrac{{236.76}}{{4}} = {59.19\text{ years}^2} $ The average seal at the zoo is 6.8 years old. The population variance is 59.19 years $^2$.